Fourier series fundamentally involve the sine and cosine functions: $$a_0+\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$$ These functions are about as non-linear as you can get. But... is the Fourier series a "linear transform"?
Thanks in advance.
If you have a linear combination of functions, the resulting Fourier series is the corresponding linear combination of the Fourier series of the functions. So yes, it is linear. The key is that it is linear in the coefficients, even though the series is not linear in $x$.
Suppose the Fourier series of a periodic function $f$ with period $2\pi$ is $$ a_0+\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right) $$ and the Fourier series of $g$ is $$ q_0+\sum_{k=1}^\infty \left(q_k \cos kx+r_k \sin kx\right). $$ Then the Fourier series of $mf+ng$, where $m$ and $n$ are constants, is $$ (ma_0+nq_0)+\sum_{k=1}^\infty \Big((ma_k + nq_k) \cos kx+(mb_k + n r_k) \sin kx\Big). $$
Therefore the tranformation whose input is $f$ and whose output is $a_0,b_1,a_1,b_2,a_2,\ldots$ is a linear transformation.
Yes, the Fourier Series is linear. The coefficients $a_k$ and $b_k$ are defined in terms of integrals and integrals are linear. For example, for constants $\lambda$ and $\mu$ and function $\mathrm{f}$ and $\mathrm{g}$ we have
$$\frac{1}{L}\int_{-L}^L \left(\lambda\mathrm{f}(x)+\mu\mathrm{g}(x)\right)\cos\left(\frac{\pi n x}{L}\right) \mathrm{d}x=$$ $$\frac{\lambda}{L}\int_{-L}^L \mathrm{f}(x)\cos\left(\frac{\pi n x}{L}\right) \mathrm{d}x+\frac{\mu}{L}\int_{-L}^L \mathrm{g}(x)\cos\left(\frac{\pi n x}{L}\right) \mathrm{d}x$$
